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Home , Refrigeration, Thermodynamics

Low & Basics

Dr. Alexander Alekseev Linde AG, Innovation Management

Main Topics

• Definitions, 1st and 2nd law of thermodynamics etc. • effects • Refrigeration processes / cycles • Helium Refrigeration

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1 Thermodynamics & Basics

• Cooling • flows from warm to objects •

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Refrigerator, Definition

 Qo  P  ???

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2 Refrigerator, 1st law Conservation of   Qo  P  Qamb

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Interfaces

Cooling capacity Qo Waste heat [capacity] Qamb P

Qo

Qamb

P

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3 Refrigerator / Water analogy

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The 1st law:

P + Qo = Qamb

Cooling Capacity: Qo = 100 W

Cooling Temperature: To = 100 K

P = 100 W  Waste Heat Qamb = 200 W

P = 1 W  Waste Heat Qamb = 101 W

Is it possible?

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4 Refrigerator, Carnot equation

 Tamb To PMIN  Qo To

Qo = 100 W

To = 100 K

Tamb = 300 K

300 K  100 K P  100W   200 W MIN 100 K

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Min. power requirements (min. power consumption)

Liquid Liquid Liquid Hydrogen Helium (LIN) (LH2) (LHe)

Cooling temperature 77.4 20.1 4.2

To, K Required cooling 100 100 100

capacity Qo, W Min. power 288 1393 7043 requirements P, W

P / Qo 2.9 13.9 70.4

All numbers calculated for ambient temperature Tamb = 300 K

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5 Exercise

An older cooling is available with:

Cooling Capacity: Qo = 300 W

Cooling Temperature: To = 4.5 K

Ambient Temperature: Tamb = 293 K = ca. 20 Cels

Revamp: 4.5 K  1.8 K

4.5 K – refrigerator:

Pmin = 300 W (293 K - 4.5 K) / 4.5 K = 19 233 W = 19.2 kW

1.8 K – refrigerator:

Pmin = 300 W (293 K - 1.8 K) / 1.8 K = 45 833 W = 45.8 kW

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COP – coefficient of performance

COP = What you get / What you pay for

What you get: Cooling capacity Qo = 300 W What you pay for: Power consumption P = 75 kW (assumpt.)

Q COP  o P

COP = Qo / P = 0.3 kW / 75 kW = 0.004 = 0.4 %

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6 Carnot efficiency (or CF, FOM)

Cooling Capacity: Qo = 300 W Power Consumption: P = 75 kW COP: 0.4%

According to Carnot equation:

 Tamb To 293K  4.2K PMIN  Qo  0.3kW 19.2kW To 4.2K  COPMAX  Qo / PMIN  0.3kW /19.2kW  0.0156 1.6%

Efficiency: e  COP /COPMAX

e  0.4% /1.6%  0.25  25%

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COOLING EFFECTS

• -Thomson expansion • Expansion in turbine

• Mixing of different fluids • Simon cooling • Peltier cooling • Vortex tube • etc.

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7 Joule Thomson Expansion (adiabatic & isenthalpic)

FlaschenventilValve Joule Thomson Effect: T 2 T T 2 T1 Temperature measurementTemperatur- 200200 bar bar meßgerät StickstoffN2

Joule-Thomson Coefficient:

Ambient temperature: 300 K  T  Umgebungstemperatur:Ambient : 1 bar T = 300 K = 27 C     1    p h const Umgebungsdruck: 1 bar

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Joule Thomson expansion (adiabatic & isenthalpic) temperature - diagram

11 1-2

H1 = H2 T 22

H3 = H4

3 , K , 3 3-4 T Temperature

44

17/04/2013 Fu17/04/2013ßzeile Low temperature refrigeration, Thermodynamics & BasicsSpecific entropy 16 sec

8 Joule Thomson Expansion, temperature - diagram

1 300 NitrogenT,h - Diagramm, N2 280 300 bar 200 bar 2 260 100 bar 50 bar 240 40 bar

220 3 200 30 bar

, K , 180 20 bar 10 bar 160 5 bar

1 bar

Temperatur, K Temperatur, 140

120 Temperature 4

100

80

60

40 7000 9000 11000 13000 15000 17000 19000 21000 Specificspezifischeenthalpy Enthalpie,, J/mol J/mol

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TURBINE

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9 Expansion in a turbine

1st law: P = H1 – H2

P = 0 

H1 = H2 (JT throttling)

P > 0 

H2 < H1  T2 < T1

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Expansion in a turbine

1 300 NitrogenT,h - Diagramm, N2 280 300 bar 200 bar 260 100 bar 50 bar 240 40 bar

220 3 200 30 bar

20 bar

, K , 180 10 bar 160 5 bar

2 1 bar Temperatur, K Temperatur, 140

120

Temperature 4 100

80

60

40 7000 9000 11000 13000 15000 17000 19000 21000 Specificspezifischeenthalpy Enthalpie,, J/mol J/mol

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10 Expansion in a turbine, T,s-Diagramm

1 1 1-2 T D

3 3 , K , 3-4 T D

2 2 Temperature

4  1 4      p      in  Pideal  M  R Tin 1     1  pout   Specific entropy  

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REFRIGERATION PROCESSES

• Joule-Thomson - process / refrigerator • Brayton - process / refrigerator • Claude - process / refrigerator

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11 Joule-Thomson process

r e o g s a s t e s

r i t p l u m o M c r e g n a h c x e e

v t l a a e v

H e l t t o r h T

Qo = H5 – H1

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Main Hardware Components:

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12 Main Hardware Components: Heat Exchanger (Alu Plate Fin Heat Exchanger)

Dimensions per core: up to 1.5 x 1.5 x 8.0 m

Standard design temperature: -269 to +65°C (150°F)

Design : up to 115 bar (1640 psig)

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Main Elements – Heat Exchanger

Modular design allows scale up to any size

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13 Main Hardware Components: Spiral Wound Heat Exchanger

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JT - process: cool - down procedure

1 amb P 5 1 5 2b 3a

2c

3b 2 4

3 2d Heat , K , o 3c Cooling 2 object Temperature

3d

4 3

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14 Liquefier / Refrigerator

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JT- refrigerator

Advantages: • Simple amb P • No moving parts in cold box  reliable 1 • Produce liquid 5

Disadvantages: • High pressure (100-200 bar) • Oil-free compressor [or with oil removal unit] • Relative high cost 2 4 • Relative high maintenance • Low efficiency 3 • Small-scale only Heat o • Cooling above 70 K only, (theoretically above 50 K only) Cooling object

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15 Open cycle JT- cooler (MMR )

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Mixed JT-cooler based on an oil-lubricated compressor

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16 Brayton-process

Qo = Ptur + H5 – H1

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Brayton-Verfahren

1. Cooling capacity Qo depends on - enthalpy difference at the warm end of heat exchanger (JT effect) - and expander [turbine] power: Qo = (H5 - H1) + Ptur

2. Working pressures used in a are < 20 bar. The enthalpy difference (H5-H1) is therefore relative small: (H5 - H1)  0

3. Therefore the cooling capacity of the Brayton cycle is defined by the power of expander [turbine]: Qo ≈ Ptur

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17 Brayton, T,s – Diagram, Nitrogen

11 5 , K ,

2

Temperature 4

3

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T,s-Diagram, Helium

10 bar

1 bar

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18 T,s – Diagram, Helium

10 bar

2

4

1 bar

3

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Brayton- refrigerator

Advantages: • High efficiency • Pressure relative low • Low cooling temperature

Disadvantages:

• Cooling temperature To = var • Liquid production: very limited • Turbine • Oilfree compressor or with oil removal unit or turbocompressor • Small cooling capacity (< 500 W) difficult to realize

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19 Brayton- Cooler

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Claude - process

Features: • Combination of an expander (turbine or expander) and JT valve (throttle) • intermediate pressure level (< 60 bar for nitrogen < 20 bar for helium) • gas at the outlet of expander • liquid at the outlet of JT valve

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20 N2- Claude- Process, T-s-diagram

1 5 5a

2a

4a 2 3a 3 4

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Claude- process

Advantages . High efficiency . Compressor pressures relative low . Liquefaction possible . Therefore stable cooling temperature . Low cooling temperatures

Disadvantages . Turbine . Oil-free compressor [or with oil removal unit] . Small cooling capacity difficult to realize

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21 Overview,

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Helium Refrigeration

• Main applications • Helium – Claude cycles • Helium Liquefier, PFD • Main Hardware Components (Heat exchangers, Compressor, Expander)

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22 Helium refrigeration, Main applications

Cooling Cooling Application temperature Cryogen Capacity Process

Bulk Helium Liquefaction 4.3 K He 1000-4000 Claude l/h Laboratory Helium Liquefier 4.3-4.6 K He 20-200 l/h Claude

High Energy 1.8K, 4.4K, He 1-20 kW Claude 80K

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N2-Claude-Verfahren He-Claude-Verfahren Pressure Pressure ratio ratio small high (>10) (<10)

single Turbine two turbines in parallel two turbines seriell

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23 Helium Liquefier, PFD

Liquid nitrogen pre-cooling

Crude-He - Purification

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1

2 4

3

LHe

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24 , overview

1000

100

turbo screw piston

Pressure, bar Pressure, 10

1 1,E+02 1,E+03 1,E+04 1,E+05 1,E+06 Flow, m3/h

Screw compressors, advantages / featues • Cost efficient, because of bulk production • Suction pressure: 1 bar (because originally air compressor) • Outlet pressure: 8-10 bar (single stage)

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Screw compressor

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25 Helium Screw Compressor Air cooled version

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Single stage compression, isentrop ideal

 1   1   p     p   T T  out  Power  M  R T   out  1 out in  p  in     in   1  pin   

AIR He R = 8,3135 J/mol/s = 1,4 1,7

M = 100 mol/s pout / pin = 9,9 9,9

Tin = 300 K Tout = 577 770 K

pin = 1,013 bar 304 497 Cels

pout = 10 bar Power = 806 949 kW

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26 He- Compression

• High temperature at the outlet (because of κ = 1,7) • High efficient cooling is necessary • Compressor oil is used not for lubrication purposes only, but for cooling purposes mainly (injected directly)

• Compressor oil should be separated  multistage oil removal (1st filter, fine filter (coalescer) and activ coal adsorber)

• Turbocompressors, Why not?

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Adiabatic expansion, ideal

1   1   p     p   T T  in  Power  M  R T  1  in   out in   in      pout   1  pout   

Luft He R = 8,3135 J/mol/s kappa = 1,4 1,7

M = 100 mol/s pin / pout = 9,9 9,9

Tin = 300 K Tout = 156 117 K

pin = 10 bar -117 -156 Cels

pout = 1,013 bar Power = -230 -391 kW

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27 Helium Expander Bearing options

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Helium Expander

Compressor impeller

Water cooler

Expander Impeller

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28 Helium Liquefier Horizontal design

— For research and develop- ment, industrial or medical application

— For the liquefaction of helium, hydrogen and neon for the production of cold

— For the recovery, purification, storage and application of helium and hydrogen

— Space

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Helium Liquefier Vertical design

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29 Helium Liquefier Infrastructure

1 - Liquefier, 2 - Compressor, 3 - Oil removal system, 4 - buffer tank, 5 – pressure control panel, 6 – main dewar vessel, 7 - mobile dewar vessel, 8 - line drier, 9 - control panel, 10 - high pressure recovery compressor, 11- high pressure vessel, 12 – gas bag

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Thank you for your attention.

Dr. A. Alekseev Linde AG, Innovation Management

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